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Algebra-II: Companion notes

Abhay Soman

Section 7.2 Polynomial ring over a UFD is a UFD

We show that a polynomial ring over a UFD is again a UFD.
Suppose that \(f\in A[x]\) is a nonconstant irreducible polynomial in \(A[x]\text{.}\) By Lemma 7.1.4, \(f\) is primitive. Suppose that \(f=gh\in Q[x]\) for some polynomials in \(g,h\in Q[x]\) with \(\deg g\lt\deg f\) and \(\deg h\lt\deg f\text{.}\) By Lemma 7.1.5, there are \(\alpha,\beta\in Q\) and primitive polynomials \(g_1,h_1\in A[x]\) such that \(g=\alpha g_1\) and \(h=\beta h_1\text{.}\) Thus \(f=(\alpha\beta)g_1h_1\text{.}\) By Gauss’s lemma (Theorem 7.1.8), \(g_1h_1\in A[x]\) is a primitive polynomial. Therefore, \(\alpha\beta\in A\) is a unit in \(A\text{.}\) We thus get the result.
First observe that an irreducible element in \(A\) remains irreducible in \(A[x]\text{.}\) Thus in order to show \(A[x]\) is a UFD we first show that any nonzero nonunit polynomial \(f\in A[x]\) can be factored into a product of irreducibles. So we assume that \(f\in A[x]\) is a nonzero nonunit. By Lemma 7.1.4, there exists a primitive polynomial \(f_1\in A[x]\) such that \(f=C(f)f_1\text{.}\) Since \(A\) is UFD, \(C(f)\) can be uniquely factored into irreducibles in \(A\text{.}\) Thus we further assume that \(f\in A[x]\) is a primitive reducible polynomial. So there are \(g,h\in A[x]\) such that \(f=gh\text{,}\) and \(\deg g\lt\deg f\) and \(\deg h\lt\deg f\text{.}\) By induction on \(\deg f\) we get \(g,h\) can be factored in \(A[x]\text{,}\) thus \(f\) can be factored in a product of irreducibles in \(A[x]\text{.}\)
Now we show that the factorization obtained above is essentially unique. Suppose that \(f\) is a nonconstant primitive polynomial in \(A[x]\) and
\begin{equation*} f=p_1p_2\cdots p_r=q_1q_2\cdots q_s. \end{equation*}
By Proposition 7.2.1 each \(p_i\) and \(q_j\) remains irreducible over the quotient field \(Q\) of \(A\text{.}\) Since \(Q[x]\) is UFD we must have \(r=s\) and there are \(a_i\in Q^\times\) such that \(p_i=a_iq_{\sigma(i)}\) for some permutation \(\sigma\in S_r\text{.}\) However, as \(p_i\) and \(q_j\) are irreducibles in \(A[x]\text{,}\) by Lemma 7.1.4, \(p_i,q_j\) are primitive elements of \(A[x]\text{.}\) Therefore, \(a_i\in A^\times\) for every \(i\) (see Corollary 7.1.6), i.e., \(p_i\sim q_{\sigma(i)}\text{.}\)
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